Posted by Becky on Sunday, October 11, 2009 at 3:52pm.
A daring ranch hand sitting on a tree limb wishes to drop vertically onto a horse galloping under the tree. The constant speed of the horse is 15.0 m/s, and the distance from the limb to the level of the saddle is 3.50 m.
(a) What must be the horizontal distance between the saddle and limb when the ranch hand makes his move?
(b) How long is he in the air?
I used the kinematic equations of motion but I can't seem to get an answer that makes sense.

physics  bobpursley, Sunday, October 11, 2009 at 5:04pm
He needs to fall 3.5m
time in air: 3.5=1/2 g t^2
now having the time in air,
distance= 15*t 
physics  Becky, Sunday, October 11, 2009 at 5:16pm
To solve for the first part, I got time = 2.89 seconds, but then it shows that I got the wrong answer?

physics  j, Friday, August 28, 2015 at 6:43pm
Nah bobpursely is definitely wrong. Anyone know how to solve this?

physics  Joshua, Saturday, August 29, 2015 at 3:34pm
Yea the solution provided is incorrect.

physics  Joshua, Saturday, August 29, 2015 at 3:43pm
The correct solution is this:
3.5/4.9 = some answer s^2
then remember
someanswer s^2 = t^2
so solve for t to get .845 seconds for time in air.